递归算法中Python列表的可变性

Question

I coded up this in-place quicksort algorithm, however the modified array is not being passed up to the parent call. I am new to Python and don't understand well the pass by value/reference, mutable/immutable stuff etc. Any explanation guidance would be great!

def quickSortPartition(a, l, r):
    if len(a) < 2:
        return a
    else:
        print a, l, r
        p = a[l]
        i = l + 1
        for j in range(l + 1, r+1):
            if a[j] < p:
                temp = a[i]
                a[i] = a[j]
                a[j] = temp
                i = i + 1
        temp = a[l]
        a[l] = a[i - 1]
        a[i - 1] = temp
        firstPartition = a[:i-1]
        pivot = a[i-1]
        secondPartition = a[i:]
        if len(a[:i-1]) > 1:
            quickSortPartition(a[:i-1], 0, len(a[:i-1])-1)
        if len(a[i:]) > 1:
            quickSortPartition(a[i:], 0, len(a[i:])-1)
        return a


lines = [3, 8, 2, 5, 1, 4, 7, 6]

# print lines
quickSorted = quickSortPartition(lines, 0, len(lines)-1)

print quickSorted

Answer 1

Basically, quickSortPartition returns a sorted list, so when you make a recursive call to quickSortPartition, make sure to capture the returned values

def quickSortPartition(a, l, r):
    if len(a) < 2:
        return a
    else:
        print a, l, r
        p = a[l]
        i = l + 1
        for j in range(l + 1, r+1):
            if a[j] < p:
                temp = a[i]
                a[i] = a[j]
                a[j] = temp
                i = i + 1
        temp = a[l]
        a[l] = a[i - 1]
        a[i - 1] = temp
        firstPartition = a[:i-1]
        pivot = a[i-1]
        secondPartition = a[i:]
        if len(a[:i-1]) > 1:
            a[:i-1] = quickSortPartition(a[:i-1], 0, len(a[:i-1])-1)
        if len(a[i:]) > 1:
            a[i:] = quickSortPartition(a[i:], 0, len(a[i:])-1)
        return a


lines = [3, 8, 2, 5, 1, 4, 7, 6]

# print lines
quickSorted = quickSortPartition(lines, 0, len(lines)-1)

print quickSorted

Output:

[3, 8, 2, 5, 1, 4, 7, 6] 0 7
[1, 2] 0 1
[5, 8, 4, 7, 6] 0 4
[8, 7, 6] 0 2
[6, 7] 0 1
[1, 2, 3, 4, 5, 6, 7, 8]

Answer 2

When you subscript a list with a range, it creates a copy of the list and return it.

So when you pass a[:i] to your function, no change will be taken into account.

When you do a[i] = 3 , it will change your list.

So you might want to change your code so that your function can take a directly as input and your index i.